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Chapter 14: Problem 80

A quantity of \(6.75 \mathrm{~g}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) was placed in a 2.00-L flask. At \(648 \mathrm{~K}\), there is \(0.0345 \mathrm{~mole}\) of \(\mathrm{SO}_{2}\) present. Calculate \(K_{\mathrm{c}}\) for the reaction $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)$$

Short Answer

Expert verified
The \(K_{c}\) for the reaction is obtained by the calculated value from step 4.

Step by step solution

01

Calculate initial moles of reactant

The initial moles (\(n_{initial_{\mathrm{SO}_{2} \mathrm{Cl}_{2}}}\)) can be calculated by the formula \(n = \frac{m}{M}\), where \(m = 6.75 g\) is mass and \( M\) is the molar mass of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) which is about 134.97 g/mol.
02

Calculate the equilibrium moles of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\)

The moles of \(\mathrm{SO}_{2}\) present at equilibrium is given as 0.0345 moles. From the balanced reaction, the production of 1 mole of \(\mathrm{SO}_{2}\) corresponds to the consumption of 1 mole of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\). Hence, the moles of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) that remain at equilibrium (\(n_{eq_{\mathrm{SO}_{2} \mathrm{Cl}_{2}}}\)) = \( n_{initial_{\mathrm{SO}_{2} \mathrm{Cl}_{2}}}\) - 0.0345.
03

Calculate the equilibrium concentrations

The equilibrium concentration (\([S]\)) of a component S can be calculated using the formula \( [S] = \frac{n_{S}}{V} \), where \(n_{S}\) is the moles of the species S and V is the volume. Using the known volume of the flask (2.00L), calculate the equilibrium concentrations of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\), \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\). Note that, as per the reaction stoichiometry, the moles (and thus the concentration) of \(\mathrm{Cl}_{2}\) is equal to that of \(\mathrm{SO}_{2}\) at equilibrium.
04

Calculate the equilibrium constant

\([K_c\) is given by the formula \([K_c = \frac{[\mathrm{products}]}{[\mathrm{reactants}]}\). In this case, \(K_c = \frac{[\mathrm{SO}_{2}][\mathrm{Cl}_{2}]}{[\mathrm{SO}_{2} \mathrm{Cl}_{2}]}\). Substitute the calculated equilibrium concentrations to find the answer.

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Most popular questions from this chapter

At \(25^{\circ} \mathrm{C}\), a mixture of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) gases are in equilibrium in a cylinder fitted with a movable piston. The concentrations are \(\left[\mathrm{NO}_{2}\right]=0.0475 \mathrm{M}\) and \(\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]=0.487 \mathrm{M} .\) The volume of the gas mixture is halved by pushing down on the piston at constant temperature. Calculate the concentrations of the gases when equilibrium is reestablished. Will the color become darker or lighter after the change? [Hint: \(K_{\mathrm{c}}\) for the dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) to \(\mathrm{NO}_{2}\) is \(4.63 \times 10^{-3} . \mathrm{N}_{2} \mathrm{O}_{4}(g)\) is colorless and \(\mathrm{NO}_{2}(g)\) has a brown color. \(]\)

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